\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{1}{y}=2\\\dfrac{6}{x}-\dfrac{2}{y}=1\end{matrix}\right.\)
\(TC:\)
\(\dfrac{1}{x}=a,\dfrac{1}{y}=b\)
\(\Rightarrow\left\{{}\begin{matrix}2a+b=2\\6a-2b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4a+2b=4\\6a-2b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a+b=2\\10b=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a+b=2\\b=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
\(\begin{cases} \dfrac{2}{x} + \dfrac{1}{y} = 2 \\ \dfrac{6}{x} - \dfrac{2}{y} = 1 \\\end{cases} (ĐK: x;y \neq 0)\)
Đặt \(\dfrac{1}{x} = u \) và \(\dfrac{1}{y} = v\) (\(u;v\neq 0\)) thì hệ đã cho trở thành
\(\begin{cases} 2u + v = 2 \\ 6u - 2v = 1 \\\end{cases}\) \(<=> \begin{cases} 4u + 2v = 4 \\ 6u - 2v = 1 \\\end{cases} <=> \begin{cases} 10u = 5 \\ 2u + v = 2 \\\end{cases} <=> \begin{cases} u = \dfrac{1}{2} \\ 2 .\dfrac{1}{2} + v = 2 \\\end{cases} <=> \begin{cases} u = \dfrac{1}{2} \\ v = 1 \\\end{cases} (T/m)\)
=> \(\begin{cases} \dfrac{1}{x} = \dfrac{1}{2} \\ \dfrac{1}{y} = \dfrac{1}{1} \\\end{cases} <=> \begin{cases} x= 2 \\ y = 1 \\\end{cases} (T/m)\)
\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{1}{y}=2\\\dfrac{6}{x}-\dfrac{2}{y}=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{2}{y}=4\\\dfrac{6}{x}-\dfrac{2}{y}=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{1}{y}=4\\\dfrac{10}{x}=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\dfrac{2}{2}+\dfrac{1}{y}=4\\x=2\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\dfrac{1}{y}=3\\x=2\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)
Vậy(2;3) là nghiệm
